Problem: The square with vertices $(-a, -a), (a, -a), (-a, a), (a, a)$ is cut by the line $y = x/2$ into congruent quadrilaterals.  The perimeter of one of these congruent quadrilaterals divided by $a$ equals what?  Express your answer in simplified radical form.
Explanation: The line $y=\frac x2$ will intersect the two vertical sides of the square, as shown below:
[asy]
real f(real x)
{

return x/2;
}

import graph;
size(6cm);
real a = 8;
pair A=(-a,a), B=(a,a), C=(a,-a), D=(-a,-a);
draw(A--B--C--D--cycle);
draw(graph(f,-11,11),Arrows);
axes(Arrows(4));
dot("$(-a,a)$",A,N);
dot("$(a,a)$",B,N);
dot("$(a,-a)$",C,S);
dot("$(-a,-a)$",D,S);
real eps=0.2;
dot((8,4)^^(-8,-4));
draw(shift((10,0))*"$2a$",(-a+eps,-a/2-.5)--(a-eps,-a/2-.5),Arrows);
draw(shift((0,10))*"$a$",(a+2*eps,-a/2)--(a+2*eps,a/2),Arrows);[/asy]
The equation of the right side of the square is $x=a,$ so we have $y= \frac x2 = \frac a2,$ which means that the intersection point with the right side of the square is $\left(a, \frac a2 \right).$ Similarly, the equation of the left side of the square is $x=-a,$ so we have $y= \frac x2 = -\frac a2,$ which means that the intersection point with the left side of the square is $\left(-a, -\frac a2 \right).$ It follows that the sides of each quadrilateral have lengths $\frac a2,$ $2a,$ $\frac{3a}2,$ and $\sqrt{a^2 + (2a)^2} = a\sqrt{5},$ by the Pythagorean theorem. Hence, the perimeter of the quadrilateral is \[\frac a2 + 2a + \frac{3a}2 + a\sqrt{5} = \left(4+\sqrt5\right)a,\]and when this is divided by $a,$ we get $\boxed{4+\sqrt{5}}.$